Equation and Inequation

Chapter 2 : Equations and Inequations2.1 Quadratic Equation 04

2.2 Roots and Coefficients 08

2.3 Formation of Equations 14

2.4 Common Roots 16

2.5 Graphs 21

2.6 Rational Functions 26

2.7 Position of Roots of a Quadratic Equation 27

2.8 Equations that can be Solved by Converting

into Quadratic 35

Solved Examples 41

Exercises 48

Previous Years’ IITJEE Questions 54

Answer Key 55

2Equations

andInequations

2010, NH Elite

Every effort has been made to avoid error or omissions in this publication. In spite of this errors may creep in.It is notified that author or seller will not be responsible for any damage or loss of action to anyone of any kindin any manner, therefrom. No part of this book may be reproduced or copied in any form or by any means(graphic, electronic or mechanical, including photocopying, recording, taping or information retrieval system)or reproduced on any disc, tape or perforated media or other information storage device, etc., without thewritten permission of NH Elite Education Services Limited. Breach of this condition is liable for legal action.

All rights reserved.

All disputes are subject to Delhi Jurisdiction only.

Corporate Office: 30 Hauz Khas Village, 3rd FloorPower House Complex, New Delhi-110016

Phone: 011- 44160000

Mathematician

Diophantus

Interesting Fact

The first known solution of a quadratic equation is

given in the Berlin papyrus from the Middle King-

dom in Egypt. This problem reduces to solving

2 2 100x y+ =

34

y x=

The Greeks were able to solve the quadratic equa-

tion by geometric methods, and Euclid's Data

contains three problems involving quadratics. Infact,

in his work Arithmetica, the Greek mathematician

Diophantus solved the above quadratic equation

by the help of Geometry, though giving only one

root.

“It is the

supreme art of

the teacher to

awaken joy in

creative

expression and

knowledge.” Albert Einstein

Diophantus of Alexandria, sometimes called "the fatherof algebra", was an Alexandrian Greek mathematicianand the author of a series of books called Arithmetica.These texts deal with solving algebraic equations, manyof which are now lost. In studying Arithmetica, Pierre deFermat concluded that a certain equation considered byDiophantus had no solutions, and noted without elabora-tion that he had found "a truly marvelous proof of thisproposition," now referred to as the very well knownFermat's Last Theorem.

This led to tremendous advances in number theory, andthe study of Diophantine equations ("Diophantine geom-etry") and of Diophantine approximations remain impor-tant areas of mathematical research. Diophantus wasthe first Greek mathematician who recognized fractionsas numbers; thus he allowed positive rational numbersfor the coefficients and solutions. In modern use, Diophan-tine equations are usually algebraic equations with inte-ger coefficients, for which integer solutions are sought.

Diophantus also made advances in mathematical nota-tion.

EQUATIONS AND INEQUATIONS IITJEE COURSE

4 |MATHEMATICS IIT STUDY CIRCLE

Section - 2.1 Quadratic Equations

2.1.1 Quadratic Equations

An equation of the form 2 0ax bx c+ + = .....(1) where 0a ≠ and , ,a b c R∈ is called a quadratic equation.

The quadratic equation 2 0ax bx c+ + = is called incomplete if at least one of the coefficients b or c is zero.

The quantity 2 4D b ac= − is known as the discriminant of the equation (1).

2.1.2 Roots of Quadratic Equation

Roots of the quadratic equation 2 0ax bx c+ + = ( 0a ≠ , , ,a b c R∈ ) are given by

2 4,

2

b b ac

a

− ± −α β =

• Sum of the roots b

a

−= α + β =

• Product of the roots c

a= α ⋅ β =

• Factorised form of ( )2 ( )ax bx c a x x+ + = − α − β .

• If S be the sum and P be the product of the roots, then the quadratic equation is : 2 0x Sx P− + = .

2.1.3 Nature of roots of a Quadratic Equation

1. Nature of quadratic equation ( )2 0 0, , ,ax bx c a a b c R+ + = ≠ ∈ means whether the roots are real or complex (no

real roots).

(a) The quadratic has real and distinct roots if and only if D > 0.

(b) The quadratic has real and equal roots if and only if D = 0.

(c) The quadratic has complex roots with non-zero imaginary parts if and only if D < 0.

2. If ( , , 0)p iq p q R q+ ∈ ≠ is a root of quadratic then p iq− is also a root of quadratic.

3. If , ,a b c Q∈ (the set of rational numbers) and D is a perfect square of a rational number, then quadratic has rational

roots.

4. If , ,a b c Q∈ and ( ), ,p q p q Q+ ∈ is an irrational root of quadratic , then p q− is also a root of quadratic .

5. If 1, ,a b c I= ∈ and the roots of quadratic are rational numbers, then these roots must be integers.

6. If quadratic is satisfied by more than two distinct roots then quadratic becomes an identity, that is 0a b c= = = .

IITJEE COURSE EQUATIONS AND INEQUATIONS

IIT STUDY CIRCLE MATHEMATICS| 5

2.1.4 Properties, related to nature of roots of Quadratic Equation

1. If 1

D and 2

D are discriminant of two quadratic equations and 1 2

0D D+ ≥ .

• At least one of 1

D and 2

0D ≥ .

• At least one of the equation has real roots.

2.1 2

0D D+ <

• At least one of 1

D and 2

0D < .

• At least one of the equation has imaginary roots.

3. If 1 2

0D D < .

• 10D > and

20D < or

10D < and

20D > .

Then one of the equations has real root and other equation has imaginary roots.

4. If 1 2

0D D >

Case I :1

0D > and 2

0D > .

Then the equations has four real roots.

Case II : 1

0D < and 2

0D < .

Then the equations has no real roots.

5. If 1 2

0D D = .

Case I :1

0D > and 2

0D = or 1

0D = and 2

0D > .

Then the equations has two same real roots and two distinct real roots.

Case II : 1

0D < and 2

0D = or 1

0D = and 2

0D < .

Then the equations has two same real roots & two imaginary roots solution.

6. If the sum of the coefficients of equation 2 0ax bx c+ + = is zero i.e., 0a b c+ + = then 1 is a root of this equation.

Illustration 1:

Comment upon the nature of the following equations :

(a) 2 2( ) 0x a b x c+ + − = (b) 2( ) 2( ) ( ) 0a b c x a b x a b c+ + − + + + − =

(c) 2( ) ( ) ( ) 0b c x c a x a b− + − + − =

Solution :

To comment the nature of quadratic equation we have to find D (Discriminant)

(a) Find Discriminant (D).

2 2 2 2( ) 4(1)( ) ( ) 4D a b c a b c= + − − = + +

⇒ 0D ≥ , hence the roots are real.

THEORY



(b) 24( ) 4( )( )D a b a b c a b c= + − + + + −

2 2 2 4[( ) ( ) ]a b a b c= + − + +

2 2 4 (2 )c c= =

⇒ 0D ≥ and also a perfect square, hence the roots are rational.

(c) 2( ) 4( )( )D c a b c a b= − − − −

2 2 22 4 4 4 4c a ac ab b ac bc= + − − + + −

2 2 2(2 ) 4 4 2c a b ab bc ac= + + − − +

2( 2 )c a b= + −

⇒ 0D ≥ and also a perfect square, hence the roots are rational.

Illustration 2:

Solve for x : ( ) ( )2 23 3

5 2 6 5 2 6 10x x− −

+ + − =

Solution :

Put ( )2 3

5 2 6x

t−

+ =

∴( )( )

2 3

5 2 6 5 2 6

5 2 6

x

t

− + − = −

⇒

2 31

5 2 6

x

t−

= −

∴ ( )2 3 1

5 2 6x

t

−

− =

then given equation becomes

110t

t+ =

⇒ 2 10 1 0t t− + =

⇒10 100 4

2t

± −=

5 2 6t = ±

Thus ( ) ( )2 3 1

5 2 6 5 2 6x − ±

± = ±

⇒ 2 3 1x − = ±

∴ 2 23 1 or 3 1x x− = − = −

⇒ 2 24 or 2x x= =

⇒ 2, 2x = ± ±



Illustration 3:

Find all roots of the equation 4 3 22 16 22 7 0x x x x+ − − + = if one root is 2 3+ .

Solution :

All coefficients are real, irrational roots well occur in conjugate pairs.

Hence another root is 2 3−

∴ Product of these roots ( )( )2 3 2 3x x= − − − +

( )22 3x= − −

2 4 1x x= − +

Dividing 4 3 22 16 22 7x x x x+ − − + by 2 4 1x x− + then the other quadratic factor is 2 6 7x x+ + .

then the given equation reduce in the form

( )( )2 24 1 6 7 0x x x x− + + + =

∴ 2 6 7 0x x+ + =

then6 36 28

2x

− ± −=

3 2= − ±

Hence roots are 2 3, 3 2± − ± .

Illustration 4:

The condition that the equation has real roots that are equal in magnitude but opposite in sign is

(a) 2 2b m= (b) 2 22b m=

(c) 2 22b m= (d) None of these

Solution : (b)

Clearly x ==== m is a root of the equation.Therefore, the other root must be –m. That is,

1 1 1 1

m m b m m b+ = +

− − + +

⇒1 1 2

b m b m m− =

− +

⇒ 2 2

2b m b m

b m m

+ − + =−

⇒ 2 2 2 2 22 2 2 or 2m b m m b= − =

Illustration 5:

Let 0, 0a b> > and 0c > . Then both the roots of the equations 2 0ax bx c+ + = can

(a) be real and negative (b) have negative real parts

(c) be rational numbers (d) None of these

THEORY



Solution : (a, b, c)

We have 2 4D b ac= − . If 0D ≥ , then the roots of the equation are given by

2

b Dx

a

− ±=

As 2 24 ( 0, 0),D b ac b a c= − < > >Q it follows that the roots of the quadratic equation are negative.

In case D < 0, then the roots of the equation are given by

2

b i Dx

a

− ± −=

which have negative real parts. (Q Both a and b are positive)

Roots will be rational numbers when D = 0.

Illustration 6:

If 2( )pr q s= + , show that at least one of the equation 2 20; 0x px q x rx s+ + = + + = has real roots.

Solution :

2 2

1 Discriminant of 0 is 4D x px q p q= + + = −

2 2

2 Discriminant of 0 is 4D x rx s r s= + + = −

Now, 2 2

1 2 4( )D D p r q s+ = + − +

also 4( ) 2q s pr+ = (Given)

Hence, 2 2 2

1 2 2 ( ) 0D D p r pr p r+ = + − = − ≥

Thus, as 1 2 0D D+ ≥ ⇒ atleast 1D or 2D must be 0≥ .

Hence, atleast one of the two equations will have real roots.

Section - 2.2 Roots and Coefficients

2.2.1 Relation between Roots and Coefficients

If 1 2, ,......,

nα α α are roots of the equation

1

1 1 0( ) ......n n

n nf x a x a x a x a−

−= + + + + ..... (1)

then 1 2

( ) ( )( )......( )n n

f x a x x x= − α − α − α

∴ 1 2

1 2 1 0 1 2...... ( )( )......( )n n n

n n n n na x a x a x a x a a x x x− −

− −+ + + + + = − α − α − α

Comparing the coefficient of like powers of x on both sides, we get

1

1 1 21

.....n

n

i ni

n

aS

a

−

== Σ α = α + α + + α =



1coefficient of

coefficient of

n

n

x

x

−

=

2 1 2 1 3.....

n

i ji j

S≠

= Σ α α = α α + α α +

2 2( 1) n

n

a

a

−= −

2

2 coefficient of ( 1)

coefficient of

n

n

x

x

−

= −

3 1 2 3 2 3 4.....

n

i j ki j k

S≠ ≠

= Σ α α α = α α α + α α α +

3 3( 1) n

n

a

a

−= −

3

3 coefficient of ( 1)

coefficient of

n

n

x

x

−

= −

M

1 2 3.....

n nS a a a a=

0constant term

( 1) ( 1)coefficient of

n n

n

n

a

a x= − = −

(A) Quadratic Equation

2 0ax bx c+ + =Let its roots be α and β

; b c

a aα + β = − αβ =

(B) Cubic Equation

3 2 0ax bx cx d+ + + = and

Let its roots be ,α β and γ

∴b

aα+β+γ = −

c

aαβ + βγ + γα =

d

aαβγ = −

THEORY



(C) Biquadratic Equation

4 3 2 0ax bx cx dx e+ + + + =Let its roots be α, β, γ and δ

b

aα + β + γ + δ = −

c

aαβ + βγ + γδ + αγ + βδ + γδ =

d

aαβγ + βγδ + γδα + αβδ = −

e

aαβγδ =

Following formulas are useful in finding the expres sion

1. 2 2 2( ) 2α + β = α + β − αβ

2.3 3 3( ) 3 ( )α + β = α + β − αβ α + β

3. (i) 4 4 3 3 2 2( )( ) ( )α + β = α + β α + β − αβ α + β

(ii) 4 4 4 2 2 2 2( ) 6 ( ) 4α + β = α + β − αβ α + β − α β

4. (i)5 5 3 3 2 2 2 2( )( ) ( )α + β = α + β α + β − α β α + β

(ii) 5 5 5 3 3 2 2( ) 5 ( ) 10 ( )α + β = α + β − αβ α + β − α β α + β

5. 2( )α − β = α + β − αβ

6.2 2 ( )( )α − β = α + β α − β

7. 3 3 2( )[( ) ]α − β = α − β α + β − αβ

8. 4 4 2 2 2 2 2 2( )( )α + β + α β = α + β + αβ α + β − αβ

9. 4 4 2 2( )( )( )α − β = α + β α − β α + β − αβ

Illustration 7:

If α and β are the roots of equations, make the coefficient of α2 common and then subtract the two equations.

(i) 2 2α + β (ii) 3 3α + β (iii) 4 4α + β

(iv) ( )2α − β (v) 4 4α − β



Solution :

In such type of problems, try to represent the given expression in terms of α + β (sum of roots) and αβ (product of roots).

In the given problem : ,b c

a a

−α + β = αβ =

(i) ( )2 2

22 2

2

2 22

b c b ac

a a a

− α + β = α + β − αβ = − − =

(ii) ( ) ( )3 3

33 3

3

33 3

b c b b abc

a a a a

− + α + β = α + β − αβ α + β = − − − =

(iii) ( ) ( ) ( )22 2 24 4 2 2 2 22 2 2 α + β = α + β − α β = α + β − αβ − αβ

( )22 2 2 22 2

2 2 4

2 222

b ac c ab ac c

a a a

− −− = − =

(iv) ( ) ( )2 2

2 2

2 2

4 44

b c b ac

a a a

−α − β = α + β − αβ = − =

(v) ( ) ( )( )24 4 2 2α −β = α + β α + β α −β

2 2

2 2

2 4b ac b b ac

a a a

− − = − ± using (i) and

D

a

α −β = ±

( )2 2

42 4

bb ac b ac

a= ± − −

Illustration 8:

If α and β are the roots of equations 2 0,ax bx c+ + = form an equation whose roots are :

(i)1 1

, +α + ββ α

(ii)1 1 1

, +α + β α β

Solution :

We know that to form an equation whose roots are known we have to find sum and product of the roots.

(i) Sum (S) ( ) ( )1 1 ( )b a c

ac

α + β − + = α + + β + = α + β + = β α αβ

Product (P) 21 1 1 ( )

2c a

ac

+ = α + β + = αβ + + = β α αβ

The equation : 2 0x Sx P− + =

⇒2

2 ( ) ( )0

b a c c ax x

ac ac

− + + − + =

⇒ 2 2( ) ( ) 0acx b c a x c a+ + + + = is the required equation.

THEORY



(ii) Sum (S) ( ) 21 1 1 1 ( )ac b

ac

α + β += + + = + = − α + β α β α + β αβ

Product (P) 1 1 1 1 a

c

= + = = α + β α β αβ


⇒2

2 ( )0

ac b ax x

bc c

+ − − + =

⇒ 2 2( ) 0bcx ac b x ab+ + + = is the required equation.

Illustration 9:

Form an equation whose roots are squares of the sum and the difference of the roots of the equation

2 2 22 2( ) 0x m n x m n+ + + + = .

Solution :

Let ,α β are the roots of the given equation

⇒ ( )m nα + β = − + and 2 2( )

2

m n+α β =

Now we have to make an equation whose roots are ( )2α + β and ( )2α − β .

Sum (S) ( ) ( ) ( ) ( )2 2 22 22 2 2 4mn = α + β + α − β = α + β = α + β − αβ =

Product ( ) ( ) ( ) ( )2 2 2 2. . 4 = α + β α − β = α + β α + β − αβ

( ) ( ) ( ) ( )22 2 2 2 2 22P m n m n m n m n = + + − + = − −


⇒ The required equation is 2 2 2 24 ( ) 0x mnx m n− − − =

Illustration 10:

If α is a root of 24 2 1 0x x+ − = . Prove that 34 3α − α is the other root.

Solution :

Let other root is β

then2 1

4 2α + β = − = −

∴ 1

2β = − − α ..... (1)

and 24 2 1 0α + α − = , because a is a root of 24 2 1 0x x+ − = .



Now ( )3 24 3 4 3α − α = α α −

(1 2 3)= α − α − { }24 2 1 0α + α − =Q

22 2= − α − α

( )214 2

2= − α − α

( )11 2 2

2= − − α − α 2{ 4 1 2 }α = − αQ

1

2= − − α

= β {from (1)}

Hence 34 3α − α is the other root.

Illustration 11:

If α, β are the roots of the equation 2( ) 5 0x x xλ − + + = . If 1λ and 2λ are two values of λ for which the roots α, β are

related by : 4

5

α β+ =β α

, find the value of 1 2

2 1

λ λ+λ λ

.

Solution :

The given equation can be written as

( )2 1 5 0x xλ − λ − + =

Q α, β are the roots of this equation

∴ 1λ −α + β =λ

and 5αβ =λ

But given 4

5

α β+ =β α

⇒2 2 4

5

α + β =αβ

⇒( )2

2 4

5

α + β − αβ=

αβ

⇒

( )2

2

1 104

5 / 5

λ −−

λ λ =λ

⇒( )2

1 10 4

5 5

λ − − λ=

λ

⇒ 2 12 1 4λ − λ + = λ

⇒ 2 16 1 0λ − λ + =

THEORY



It is quadratic in λ, let roots be 1λ and 2λ then

1 2 16λ + λ = and 1 2 1λ λ =

∴2 2

1 2 1 2

2 1 1 2

λ λ λ + λ+ =λ λ λ λ

( )2

1 2 1 2

1 2

2λ + λ − λ λ=

λ λ

2(16) 2

1

−=

= 256 – 2

= 254

Section - 2.3 Formation of Equations

We now list some of the methods to form an equation whose roots are given in terms of some other equation.

Let the given equation be :

1 2

0 1 2 1..... 0n n n

n na x a x a x a x a− −

−+ + + + + = ..... (1)

Methods to Transform the Equations

Method 1 : To form an equation whose roots are ( 0)k ≠ times roots of the equation in (1), replace x by d/k in (1).

Method 2 : To form an equation whose roots are the negative of the roots in (1) replace x by –x in (1). Alternatively change

the sign of the coefficients of 1 3 5, , ,......n n nx x x− − −

etc. in (1).

Method 3 : To form an equation whose roots are k more than the roots in (1) replace x by x – k in (1).

Method 4 : To form an equation whose roots are reciprocals of the roots in equation (1) replace x by 1

x in (1) and then

multiply both the sides by nx .

Method 5 : To form an equation whose roots are square of the roots of the equation (1), proceed as follows :

Step 1 : Replace x by x in (1).

Step 2 : Collect all the terms involving x on one side.

Step 3 : Square both the sides and simplify.

Method 6 : To form an equation whose roots are the cubes of the roots of the equation (1) proceed as follows :

Step 1 : Replace x by 1/3x .

Step 2 : Collect all the terms involving 1/3x on one side.

Step 3 : Cube both the sides and simplify.



Illustration 12:

Let α, β are the roots of the quadratic 2 0ax bx c+ + = . Find the quadratic whose roots are

(a) 2 ,2α β (b) ,−α −β (c) 1, 1α + β +

(d)1 1

,α β

(e) 2 2,α β

Solution :

(a) 2 0ax bx c+ + = has roots α, β.

( )2/2 ( /2) 0a x b x c+ + = has roots 2α, 2β.

⇒ 2 2 4 0ax bx c+ + =

(b) 2 ( ) 0ax bx c f x+ + = = has roots α, β.

so, ( ) 0f x− = has roots ,− α −β .

( ) ( )20a x b x c− + − + = has roots ,− α −β .

⇒ 2 0ax bx c− + =2 0ax bx c− + =

(c) 2( ) 0f x ax bx c= + + = has roots α, β.

( 1) 0f x − = has roots 1, 1α + β + .

2( 1) ( 1) 0a x b x c− − − + = has roots 1, 1α + β + .

⇒ 2 (2 ) ( ) 0ax a b x a b c− + + + + =

(d) 2( ) 0f x ax bx c= + + = has roots α, β.

10f

x =

has roots 1 1

,α β

. (provided 0α ≠ & 0β ≠ )

21 1

0a b cx x

+ + =

⇒ 2 0cx bx a+ + =

(e) 2( ) 0f x ax bx c= + + = has roots α, β.

( ) 0f x = has roots 2 2,α β

( ) ( )2

0a x b x c+ + =

⇒ 0ax c b x+ + =

⇒ ( ) ( )22ax c b x+ =

⇒ 2 2 2 2(2 ) 0a x ac b x c+ − + =

THEORY



Section - 2.4 Common Roots

Suppose the quadratic equations

2

1 1 10a x b x c+ + = ...... (i) and

2

2 2 20a x b x c+ + = ...... (ii) where

1 20, 0a a≠ ≠ and

1 2 2 10a b a b− ≠ .

2.4.1 One Common Root

Let the common root be α

∴ 2

1 1 10a b cα + α + = ...... (iii) and

2

2 2 20a b cα + α + = ...... (iv)

solving (iii) and (iv), we get

1 2 2 1

1 2 2 1

c a c a

a b a b

−α =

−

Required condition is

2

1 2 2 1 1 2 2 1 1 2 2 1( )( ) ( )a b a b b c b c c a c a− − = −

How to obtain a common root :

The equations same and subtract one equation from the other to obtain a linear equation in x. Solve it for x to obtain the commonroot.

2.4.2 Both roots are common

If two quadratic has both roots in common, then their coefficients are in proportion.

Condition : If 21 1 1 0a x b x c+ + = and 2

2 2 2 0a x b x c+ + = have both roots in common then

1 1 1

2 2 2

a b ca b c

= =

Illustration 13:

Find the value of k, so that the equations 22 5 0x kx+ − = and 2 3 4 0x x− − = may have one root in common.

Solution :

Let α be the common root of two equations.

Hence 22 5 0kxα + − = and 2 3 4 0α − α − =Solving the two equations :

2 1

4 15 8 5 6k k

α −α= =− − − + − −

⇒ 2( 3) (4 15)(6 )k k− = + +

⇒ 24 39 81 0k k+ + =

⇒ 3 or 27/4k k= − = −



Illustration 14:

If the equations 2 0x ax b− + = and 2 0x cx d− + = have one root in common and second equation has equal roots, prove

that 2( )ac b d= + .

Solution :

The equation 2 0x cx d− + = has equal roots.

⇒ 0D = ⇒ 2 4 0D c d= − = ...... (i)

Also, the equal roots are 2

bx

a= − (for 2 0ax bx c+ + = having equal roots)

⇒2

cx = is the equal root of this equation.

Now this should be the common root.

∴2

cx = will satisfy the first equation.

⇒2

04 2

c ca b − + =

⇒ 2 4 2c b ac+ =

⇒ 4 4 2d b ac+ = using (i)

⇒ 2( )d b ac+ =

Hence 2( )ac b d= +

Illustration 15:

If α, β are the roots of 2 0x px q+ + = and γ, δ are the roots of 2 0x rx s+ + = , evaluate the value of ( )( )α − γ α − δ

( )( )β − γ β − δ in terms of p, q, r, s. Hence deduce the condition that the equations have a common root.

Solution :

Let α, β be the roots of 2 0x px q+ + =

⇒ pα + β = − and qαβ = ..... (i)

γ, δ are the roots of 2 0x rx s+ + =

⇒ rγ + δ = − and sγ δ = ..... (ii)

Expanding ( )( )( )( )α − γ α − δ β − γ β − δ

( ) ( )2 2= α − γ + δ α + γδ β − γ + δ β + γδ [using (i) and (ii)]

( )( )2 2r s r s= α + α + β + β +

As α is a root of 2 0x px q+ + = ,

We have 2 0p qα + α + =

THEORY



And similarly 2 0p qβ + β + =

Substituting the values of 2α and 2β , we get ;

( )( )( )( )α − γ α − δ β − γ β − δ

( )( )p q r s p q r s= − α − + α + − β − + β +

( ) ( )r p s q r p s q= − α + − − β + −

( ) ( ) ( )( )( )2 2r p s q s q r p= − αβ + − + − − α + β

( ) ( ) ( )( )2 2r p q s q p s q r p= − + − + − −

( )( ) ( )2r p rq pq ps pq s q= − − − + + −

( )( ) ( )2r p qr ps s q= − − + −

If the equations have a common root then either

or or or α = γ α = δ β = γ β = δ

i.e., ( )( )( )( ) 0α − γ α − δ β − γ β − δ =

⇒ ( ) ( )( )20s q r p qr ps− + − − =

⇒ ( ) ( )( )2s q r p qr ps− = − −

2.4.3 Imaginary Roots

If the two equations 2

1 1 10a x b x c+ + = and 2

2 2 20a x b x c+ + = with real coefficients have an imaginary root common,

then both roots will be common.

Condition : 1 1 1

2 2 2

a b c

a b c= =

2.4.4 Irrational Roots

If the two equations 2

1 1 10a x b x c+ + = and 2

2 2 20a x b x c+ + = with rational coefficients have an irrational root common,

the roots will be common.

Condition : 1 1 1

2 2 2

a b c

a b c= =

Illustration 16:

Find all the roots of the 4 3 24 24 57 18 45 0x x x x− + + − = equation 4 3 24 24 57 18 45 0x x x x− + + − = if one root is

3 6i+ .



Solution :

As the coefficients are real, complex roots will occur in conjugate pairs. So the other root is 3 6i− .

Let α, β be the remaining roots.

⇒ the four roots are 3 6,i± α and β

Hence the factors are :

( )( )( )( )3 6 3 6x i x i x x= − − − − − α − β

( ) ( )( )23 6x x x = − + − α − β

( )( )( )2 6 15x x x x= − + − α − β

Dividing 4 3 24 24 57 18 45x x x x− + + − by 2 6 15x x− + or by inspection we can find that the other factor of given equation

is 24 3x − .

⇒ ( )( )4 3 2 2 24 24 57 18 45 6 15 4 3x x x x x x x− + + − = − + −

⇒ α, β are roots of 24 3 0x − =

⇒ , 3 / 2α β = ± Hence roots are 3 6, 3 / 2i± ±

Illustration 17:

If equations 2 0ax bx c+ + = and 2 2 3 0x x+ + = have a common root, show that : : 1: 2 : 3a b c= .

Solution :

Since roots of 2 2 3 0x x+ + = are imaginary and equations 2 2 3 0x x+ + = & 2 0ax bx c+ + = have a common root,therefore both roots will be common. Hence both equations are identical.

∴1 2 3

a b c= =

∴ : : 1: 2 : 3a b c=

Illustration 18:

Find the minimum value of ( )m n p+ + where , ,m n p +∈� . Given that 2 0mx nx p+ + = and 2 4 1 0x x+ + = has a

common root.

Solution :

Roots of 2 4 1 0x x+ + = are 2 3− ± .

And irrational roots occur in conjugate pair.

Hence, the equations have both not in common.

∴1 2 3

a b c= =

So,1 4 1

m n p= =

4 6m n p+ + = λ + λ + λ = λ , where λ ∈� .

THEORY



for minimum value of ( )m n p+ + , 1λ = .

so minimum value of ( )m n p+ + is 6.

Illustration 19:

The values of 'a' for which the equations 3 26 (6 ) 6 0x x a x− + + − = and 2 4 0x ax− + = have a common root.

Solution :

Let 'α' is the common root.

3 2

2

6 (6 ) 6 0

4 0

a

a

α − α + + α − =α − α + =

Adding 3 25 6 2 0α − α + α − =

⇒ 1α = is one solution of above equation.

3 2

3 2

2

2

2 4 2

5 6 2

1 4 6 2

4 4

2 2

α − α+

α − α + α −α − α

α − − α + α −− α + α

α −

⇒ ( )( )21 4 2 0α − α − α + =

⇒ ( ) ( ){ } ( ){ }1 2 2 2 2 0α − α − − α − + =

1, 2 2 or 2 2α = α = − α = +

This common root can be 1, 2 2− , 2 2+ .

We can obtain values of 'a'.

1α = ⇒ 1 4 0 5a a− + = ⇒ =

2 2α = − ⇒ ( ) ( )6 4 2 2 2 4 0a− − − + =

⇒ ( ) ( )10 4 2 2 2a− = −

⇒( )( )

( )10 4 2 2 210 4 2

4 22 2a

− +−= =−−

( )( )5 2 2 2 2a = − +

10 5 2 4 2 4a = + − −

6 2a = +

( ) ( )2 2 6 4 2 2 2 4 0aα = + ⇒ + − + + =



⇒( )( )10 4 2 2 210 4 2

22 2a

+ −+= =+

⇒ ( )( )5 2 2 2 2a = + −

10 5 2 4 2 4a = − + −

6 2a = −

Thus possible values of 'a' are 1, 6 2±

In-Chapter Exercise - 1

1. If the roots of the quadratic equation 2

34 log 0x x a− − = are real the least value of 'a' is :

(a) 81 (b) 1/81 (c) 1/64 (d) none of these

2. If x is an integer satisfying 2 6 5 0x x− + ≤ and 2 2 0x x− > then the number of possible integral values of x is:

(a) 3 (b) 4 (c) 2 (d) Infinite

3. If a, b, c are non zero unequal rational numbers then the roots of the equation

2 2 2 2 2 2(3 ) 6 2 0abc x a b cx a ab b+ + − − + = are :

(a) rational (b) imaginary (c) irrational (d) none of these

4. For x R∈ , prove that the given expression 2

2

34 71

2 7

x x

x x

+ −+ −

can not lie between 5 and 9 ?

Section - 2.5 Graphs

2.5.1 Quadratic Expression and its Graphs

In general quadratic expression is represented as

2( )f x ax bx c= + + or 2 , 0, , , Ry ax bx c a a b c= + + ≠ ∈

2( )b c

f x a x xa a

= + +

2 2

2

4

2 4

b ac ba x

a a

−= + +

2

22 4

b Da x

a a= + −

THEORY



2

2 4

b Dy a x

a a= + −

⇒

21

2 4

b Dx y

a a a+ = +

To draw the graph of ( )f x , proceed according to following steps.

Step - I : Shape of Curve :

The shape of the curve ( )y f x= is parabolic with following conditions.

For a > 0, the parabola opens upwards.

For a < 0, the parabola opens downwards.

a > 0

a < 0

x-axis

Step - II : Intersection with axes :

(i) With x-axis

For D > 0, parabola cuts x-axis in two points. aD > 0 > 0

a D > 0, > 0

The points of intersection are α, β

2

b D

a

±=

(ii) For D = 0

Parabola touches x-axis in one point. a D > 0, = 0

a D < 0, = 0

The point of intersection is 2

b

aα = −

(iii) For D < 0

Parabola does not cut x-axis at all i.e., no point of intersection with x-axis.

a D > 0, < 0

a D < 0, < 0



(iv) With y-axis

The point of intersection with y-axis is (0, 0), {Put x = 0 in the quadratic polynomial}

Step - III : Maximum and Minimum Value of 2( )f x ax bx c= + += + += + += + +

V is called the vertex of parabola.

The coordinates of V, for the quadratic 2y ax bx c= + + is ,2 4b Da a

− −

The line passing through the vertex and parallel to the y-axis is called as axis of symmetry.

O

axis of symmetry

a D < 0 and < 0

vertex

y

( ,0)β ( ,0)α O

axis of symmetry

a D > 0 and > 0

vertex

y

( ,0)β ( ,0)α

⇒ ( )f x has minimum value at vertex if 0a > and min

4

Df

a= − at

2

bx

a= − .

max . (not defined)f = ∞ .

⇒ ( )f x has maximum value at vertex if 0a < and max4

Df

a= − at

2

bx

a= − .

min . (not defined)f = −∞ .

Illustration 20:

Show that the graph of the polynomial : 2 9y x kx x= + − + is above X-axis, if and only if 5 7k− < < .

Solution :

Let 'α' is the common root.

2y ax bx c= + + has its graph above x-axis if : a > 0 and D < 0

Given ( )2 1 9y x k x= + − +

Coefficient of 2 1x a= = i.e., positive–5 7k

( )21 36 0D k= − − < , for graph to lie above x-axis.

( )( )7 5 0k k− + <

⇒ 5 7k− < < {For graph to lie above x-axis}

THEORY



Illustration 21:

Find f (max) or f (min) in the following polynomials over x R∈ .

(i) 2( ) 4 12 15f x x x= − + (ii) 2( ) 3 5 4f x x x= − + −

Solution :

(i) 2( ) 4 12 15f x x x= − +

As 4 0a = >

( )f x has minimum value at vertex.

2( 12) 4 4 15 144 240 96D = − − × × = − = −

min 4

Df

a

−= at 2

bx

a

−=

⇒ min

( 96) 966

4 4 16f

− −= = =×

at 12 3

2 4 2x

−= =×

∴min 6f = at

3

2x =

maxf = ∞

(ii) 2( ) 3 5 4f x x x= − + −

As 3 0a = − <

( )f x has maximum value at vertex.

2( 5) 4( 3)( 4) 25 48 23D = − − − − = − = −

max 4

Df

a

−= at 2

bx

a= −

max

( 96) 23

4( 3) 12f

−= − = −−

at (5) 5

2( 3) 6x

−= =−

∴ max

23

12f

−= at 5

6x =

minf = −∞

Illustration 22:

If ( )2

2log 1 ax x a x R+ + ≥ ∀ ∈ , then exhaustive set of values of 'a' is :

(a)5

0,12

+

(b)

5 51 ,1

2 2

− +

(c)

50,1

2

−

(d)

51 ,

2

+ ∞

Solution :

(d) ( )2

2log 1 ax x a x R+ + ≥ ∀ ∈

⇒ 2 2ax x a x R+ + ≥ ∀ ∈



⇒ ( )2 2 0ax x a x R+ + − ≥ ∀ ∈

⇒ coefficient of 2 0x > and 0D ≤

⇒ 0,1 4 ( 2) 0a a a> − − ≤

⇒ 20,4 8 1 0a a a> − − ≥

⇒5 5 5

0, ,1 1 , 1 ,2 2 2

a a a

> ∈ −∞ − ∪ + ∞ ⇒ ∈ + ∞

Sign of a Quadratic Polynomial

Let 2( ) , , , and 0f x ax bx c a b c R a= + + ∈ ≠ .

Situation Graphical Representation Results

0, 0a D> <

0, 0a D< <

0, 0a D> >

0, 0a D< >

0, 0a D> =

0, 0a D> <x

( ) 0,

for all

f x

x R

>∈

0, 0a D< <x

( ) 0,

for all

f x

x R

<∈

0, 0a D> >

xα β

( ) 0,

for all ( , ] [ , )

and ( ) 0 for all ( , )

f x

x

f x x

≥∈ −∞ α ∪ β ∞< ∈ α β

α < β

0, 0a D< >x

α β

α < β ( ) 0 for all

( , ] [ , )

and ( ) 0 for all ( , )

f x

x

f x x

≤∈ −∞ α ∪ β ∞

> ∈ α β

x

α = β0, 0a D> = ( ) 0 for

all

f x

x R

≥∈

THEORY



Section - 2.6 Rational Functions

2.6.1 Introduction to Rational Functions

The ratio of two polynomials of x say ( )P x and ( )Q x where ( ) 0Q x ≠ is called a rational function :

If( )

( ) ; ( ) 0( )

P xf x Q x

Q x= ≠

then ( )f x is a rational function of x.

Example :

(a)2

2

2, 2, 3

5 6

x xx

x x

+ +≠ ≠

− +(b) 2

1

3 4

x

x x

+

+ +

2.6.2 Maximum and Minimum Value of a Rational Func tion

Consider 2

2( )

ax bx cf x y

px qx r

+ += =

+ +; where x R∈

⇒2( ) ( ) ( ) 0a py x b qy x c ry− + − + − =

As x is real, 0D ≥

⇒2( ) 4( )( ) 0b qy a py c ry− − − − ≥ ..... (1)

This relationship is an inequality in y.

On solving the inequality we will get the values of y can take.

Case I : If [ ],y A B∈

max min,y B y A= =

Case II : If ( , ] [ , )y A B∈ −∞ ∪ ∞

maxy = ∞ i.e., not defined

miny = −∞ i.e., not defined

Case III: ( , )y ∈ −∞ ∞ , i.e., y R∈

maxy = ∞ i.e., not defined

miny = −∞ i.e., not defined

Section - 2.7 Position of Roots of a Quadratic Equation

Let 2( ) ,f x ax bx c= + + where , ,a b c R∈ be a quadratic expression and 1 2

, ,k k k be real numbers such that 1 2

k k< . Let ,α β be

the roots of the equation ( ) 0f x = .

i.e., 2 0ax bx c+ + = , then

2

b D

a

− −α = and

2

b D

a

− +β =



where D is the discriminant of the equation.

Situation Graphical Representation Required Condition

kα < β <

k < α < β

a > 0

y

α β k

,2 4

b D

a a− −

x

a < 0

α βk

x

2

b

a−

/2 , /4b a D a− −

a > 0y

α βkx

a < 0

α βk

x–b a/2–b a/2

y

( ) 0

( ) ( ) 0

( ) /2

i D

ii af k

iii k b a

≥>

> −

( ) 0

( ) ( ) 0

( ) /2

i D

ii af k

iii k b a

≥>

< −

Some More Results on Roots of Quadratic Equations

(a) Both roots of ( ) 0f x = are negative

(i) Sum of the roots < 0

Product of the roots > 0

and 0D ≥

i.e., 20, 0, 4 0b c

b aca a

− < > − ≥

(b) Both roots of ( ) 0f x = are positive

(i) Sum of the roots > 0

Product of the roots > 0

and 0D ≥

i.e., 20, 0, 4 0

b cb ac

a a− > > − ≥

(c) Roots of ( ) 0f x = are of opposite in sign.

Product of the roots < 0, i.e., 0c

a<

THEORY



kα < < β

1 2k k< α < β <

1 2k kα < < < β

1 2k k< α < < β

a < 0

xα βk

ya > 0

y

α β xk

a > 0

α βx

k1 k2

a < 0

α β

x

k1 k2

a > 0

α βx

k1 k2

–b a/2

a < 0

α βx

k1 k2

–b a/2

a > 0

βx

k1

k2

a < 0

αβ

xk1

k2α

( ) 0

( ) ( ) 0

i D

ii af k

>>

1 2

1 2

0

( ) 0, ( ) 0

/2

D

af k af k

k b a k

≥> >

< − <

1

2

0

( ) 0

( ) 0

D

af k

af k

>><

1 2

0

( ) ( ) 0

D

f k f k

>× <

Illustration 23:

The least integral value of 'k ' for which 2( 2) 8 4 0k x x k− + + + > for all ,x R∈ is :

(a) 5 (b) 2 (c) 3 (d) none of these

Solution : (a)

( )2

2log 1 ax x a x R+ + ≥ ∀ ∈

Let 2( ) ( 2) 8 4f x k x x k= − + + +

( ) 0 0 and 0f x a D> ⇒ > <

2 0k − > and 64 4( 2)( 4) 0k k− − + <



2k > and 216 ( 2 8) 0k k− + − <

2k > and 2 2 24 0k k+ − >

2k > and ( 6 or 4)k k< − >

As it can be observe that k can take values greater than 4.

⇒ 4k > .

least integral value of 5k = .

Illustration 24:

If a < b, then solution of 2 ( ) 0x a b x ab+ + + < is given by

(a) or x b x a< < (b) a x b< < (c) or x a x b< > (d) b x a− < < −

Solution : (d)

2 ( ) 0x a b x ab+ + + <

⇒ ( )( ) 0x a x b+ + < ⇒ b x a− < < −

Illustration 25:

Let , ,a b c R∈ be such that 0, 0a b c a b c+ + < − + < and c > 0. If 'α' and 'β' are roots of the equation 2 0,ax bx c+ + =

then value of [ ] [ ]α + β is (where [·] denotes greatest integer of x)

(a) 2 (b) 1 (c) –1 (d) 0

Solution : (c)

Let 2( ) 0f x ax bx c= + + =

Now as 0a b c+ + < ⇒ (1) 0f < ..... (i)

0a b c− + < ⇒ ( 1) 0f − < ..... (ii)

0c > ⇒ (0) 0f > ..... (iii)

combining (i), (ii) and (iii), to get

( 1) (0) 0f f− < and (0) (1) 0f f <Hence one of the root lies between (–1,0) and other root lies between (0, 1).

⇒ [ ] [ ] 1 0 1α + β = − + = − .

Illustration 26:

The adjoining figure shows the graph of 2y ax bx c= + + . Then

(a) 0a < (b) 2 4b ac>

0( ,0)x1 ( ,0)x2

x

y

(c) 0c > (d) a and b are of opposite signs

Solution : (a) & (d)

As it is clear from the figure that it is a parabola opens downwards

i.e. a < 0. ⇒ (c) is correct.

THEORY



⇒ It is 2y ax bx c= + + i.e. degree two polynomial.

Now, if 2 0 ax bx c+ + = ⇒ It has two roots 1x and 2x as it cuts the axis at two distinct point 1x and 2x .

Now from the figure it is also clear that 1 2 0x x+ > . (i.e. sum of roots are positive)

⇒ 0 0 b b

a a

− > ⇒ < ⇒ a and b are of opposite signs. ⇒ (d) is correct.

As D > 0 and (0) 0f c= < , both (b) and (c) are wrong.

Illustration 27:

The diagrams shows the graph of 2y ax bx c= + + . Then

(a) 0a > (b) 0b <

(c) 0c > (d) 2 4 0b ac− = 0( ,0)x1 ( ,0)x2

x

y

x'

Solution : (b) & (c)

As it is clear from the figure that it is a parabola opens downwards i.e. a < 0.

⇒ It is 2y ax bx c= + + i.e. degree two polynomial.

Now if 2 0 ax bx c+ + = ⇒ it has two roots 1x and 2x as it cuts the axis at two distinct point 1x and 2x .

Now from the figure it is also clear that 1 2 0x x+ < (i.e., sum of roots are negative)

⇒ 0 0 0 b b

ba a

− < ⇒ > ⇒ < ⇒ (b) is correct.

As the graph of ( )y f x= cuts the +y-axis at (0, c) where c > 0 ⇒ (c) is correct.

Illustration 28:

Prove that for all real x, y the value of 2 22 3 6 2x xy y x y+ + − − cannot be less than –11.

Solution :

We have to prove that 2 22 3 6 2 11x xy y x y+ + − − ≥ − ..... (i)

i.e., 2 22 3 6 2 11 0x xy y x y+ + − − + ≥ for all ,x y R∈ .

or 2 22( 3) 3 2 11 0x y x y y+ − + − + ≥ for all ,x y R∈ .

This is true if 0D ≤ and coefficient of x2 is positive.

Clearly coefficient of 2 1 0x = >

Now 0D ≤

Illustration 29:

Find the values of a for which one root of equation 2( 5) 2 4 0a x ax a− − + − = is smaller than 1 and the other greater than

2.



Solution :

Let ( )2( ) ( 5) 2 4 5f x a x ax a a= − − + − ≠ as 1 and 2 lie between the roots of ( ) 0f x = , we can take 0D > , ( 5) (1) 0a f− <

and ( 5) (2) 0a f− < .

I. Consider D > 0 : 2( 2 ) 4.( 5).( 4) 0a a a− − − − >

⇒ 9 20 0 (20/9, )a a− > ⇒ ∈ ∞ ..... (1)

II. Consider (a – 5) f (1) < 0 : ( )( )5 5 2 4 0a a a a− − − + − <

⇒ ( )( )5 9 0a− − <

⇒ 5 0a − >

⇒ (5, )a∈ ∞ ..... (2)

III. Consider (a – 5) f (2) < 0 : ( ) ( )5 4 5 4 4 0a a a a− − − + − <

⇒ ( )( )5 24 0a a− − <

⇒ (5,24)a∈ ..... (3)

Hence, the values of a satisfying (1), (2) and (3) at the same time are (5,24)a∈ .

Illustration 30:

For x R∈ , prove that the given expression 2

2

34 71

2 7

x x

x x

+ −+ −

can not lie between 5 and 9?

Solution :

Let 2

2

34 71

2 7

x xy

x x

+ − =+ −

⇒ 2( 1) (2 34) 71 7 0x y y x y− + − + − =

For real values of x, discriminant 0≥

∴ ( ) ( )( )22 34 4 1 71 7 0y y y− − − − ≥

⇒ 28 112 360 0y y− + ≥

⇒ 2 14 45 0y y− + ≥

⇒ ( )( )9 5 0y y− − ≥

⇒ ( ] [ ),5 9,y∈ −∞ ∪ ∞

Hence, y can never lie between 5 and 9.

Illustration 31:

Find the values of m so that the inequality : 2

2

13

1

x mx

x x

+ + <+ +

holds for all x R∈ .

THEORY



Solution :

We know that

| |a b b a b< ⇒ − < <

Hence2

2

13

1

x mx

x x

+ + <+ +

⇒2

2

13 3

1

x mx

x x

+ +− < <+ +

Case I :2

2

13

1

x mx

x x

+ + <+ +

⇒( ) ( )2 2 2

22

1 3 1 2 ( 3) 20 0

1 1 32 4

x mx x x x m x

x xx

+ + − + + − + − −< ⇒ <+ + + +

Multiplying both sides by denominator, we get :

⇒ 22 ( 3) 2 0x m x− + − − < (because denominator is always positive)

⇒ 22 ( 3) 2 0x m x− − + >

A quadratic expression in x is always positive if coefficient of 2 0x > and D < 0.

⇒ ( ) ( )( ) ( ) ( )( ) ( )2 2 23 4 2 2 0 3 4 0 7 1 0 1,7m m m m m− − < ⇒ − − < ⇒ − + < ⇒ ∈ − ..... (1)

Case II :2

2

13

1

x mx

x x

+ +− <+ +

⇒( ) ( )2 2

2

2

1 3 10 4 ( 3) 4 0

1

x mx x xx m x

x x

+ + + + +< ⇒ + + + >

+ +

For this to be true for all , 0x R D∈ <

⇒ ( ) ( )( )23 4 4 4 0m+ − <

⇒ ( )( )3 8 3 8 0m m+ − + + < [using 2 2 ( )( )a b a b a b− = + − ]

⇒ ( 5)( 11) 0m m− + <

⇒ ( 11,5)m∈ − ..... (2)

We will combine (1) and (2) because both must be satisfied.

⇒ The common solution is ( 1,5)m∈ − .

Illustration 32:

Find all the values of p for which the roots of the equation 2( 3) 2 5 0p x px p− − + = are real and positive.

Solution :

The roots are real and positive if 0,D ≥ sum of the roots > 0 and product of the roots > 0.

0D ≥ :



⇒ 2 2 24 20 ( 3) 0 4 15 0 4 15 0 [0,15/4]p p p p p p p p− − ≥ ⇒ − + ≥ ⇒ − ≤ ⇒ ∈ ..... (1)

Sum of the roots > 0 :

20 0

3 3

p p

p p> ⇒ >

− −

⇒ ( 3) 0 ( ,0) (3, )p p p− > ⇒ ∈ −∞ ∪ ∞ ..... (2)

Product of the roots > 0 :

50 0 ( 3) 0 ( ,0) (3, )

3 3

p pp p p

p p> ⇒ > ⇒ − > ⇒ ∈ −∞ ∪ ∞

− − ..... (3)

Combining (1), (2) and (3) on the number line, we get :

(3,15/4]p∈

Illustration 33:

The values of 'a' for which both the roots of 2 24 2 3 5 0x ax a a− + − + = are greater than 2, are :

(a) (1, )a∈ ∞ (b) 1a = (c) ( ,1)a∈ −∞ (d) (9/2, )a∈ ∞

Solution :

Let 2 2( ) 4 2 3 5f x x ax a a= − + − + . The conditions for both the roots to exceed 2 are

(i) 0D ≥ (ii) (2) 0f > and (iii) 2vx >

Now consider 0D ≥

⇒ 2 2 216 4(2 3 5) 0 2 3 5 0a a a a a− − + ≥ ⇒ + − ≥

⇒ (2 5)( 1) 0a a+ − ≥

⇒ ( , 5/2] [1, )a∈ −∞ − ∪ ∞

⇒ a R∈ ..... (1)

Now consider (2) 0f >

⇒ 2 24 8 (2 3 5) 0 2 11 9 0a a a a a− + − + > ⇒ − + >

⇒ (2 9)( 1) 0a a− − >

⇒ ( ,1) (9/2, )a∈ −∞ ∪ ∞ ..... (2)

Now consider 2vx >

⇒4

22

a >

⇒ a > 1 ..... (3)

on combining (1), (2) and (3), we get

(9/2, )a∈ ∞

THEORY



Some Important Properties

Let ( ) 0f x = be a quadratic equation.

(a) If ( )f a and ( )f b have opposite signs, then ( ) 0f x = has one root between a and b and other root

outside the interval [a, b], where a < b.

(b) If ( )f a and ( )f b have same signs and roots of ( )f x are real, then ( ) 0f x = has either both roots

between a and b or both roots outside the interval [a, b].

Let ( ) 0f x = be a polynomial equation.

(a) If ( )f a and ( )f b have opposite signs, then equation ( ) 0f x = will have odd number of roots in a and b.

(b) If ( )f a and ( )f b have same sign then ( ) 0f x = will have even number of roots between a and b (this also

includes the case of zero number of roots between a and b).

Illustration 34:

If the expression 2 2 2 2 2 2ax by cz ayz bzx cxy+ + + + + can be resolved into rational factors, prove that

3 3 3 3a b c abc+ + = .

Solution :

The given expression is 2 2 2 2 2 2ax by cz ayz bzx cxy+ + + + +

2 2

2 2 2 2x y y x x y

z a b c a b cz z z z z z

= + + + + +

2 2 2[ 2 2 2 ]z aX bY cXY bX aY c= + + + + + ..... (1)

where, x

Xz

= and y

Yz

= .

The given expression can be resolved into rational factors if the expression within brackets in (1) is expressible intorational factors, the condition for which is

2 2 2 3 3 32 . . . 0 3abc abc a a b b c c a b c abc+ − − − = ⇒ + + =

Section - 2.8 Equations that can be solved by converting into Qua dratic

2.8.1 Equations of Degree ‘4’

� 4 2 0ax bx c+ + =

2Put x y=

eg. : 4 23 4 0x x− − = , Let 2x y=



we get 2 3 4 0y y− − =

⇒ ( )( )4 1 0y y− + =

⇒ 2 4x = ⇒ 2 4 0x − = ⇒ 2x = ±

2 1x = − ⇒ 2 1 0x + = ⇒ ix = ±

Thus 4 23 4 0x x− − = has four roots i+ , i− , 2, –2.

�4 3 2 0+ + + + =ax bx cx bx a

2Divide by x

eg. : 4 3 22 11 2 0x x x x+ − + + =

dividing by 2x (as 0x ≠ ) ......... (why?)

22

1 22 11 0x x

x x+ − + + =

⇒ 22

1 12 11 0x x

xx + + + − =

⇒2

1 12 2 11 0x x

x x

+ − + + − =

⇒

21 1

2 15 0x xx x

+ + + − =

Put 1

x yx

+ =

we get 22 15 0y y+ − =

⇒ 22 6 5 15 0y y y+ − − =

⇒ ( )( )2 5 3 0y y− + =

5, 3

2y = −

1 5

2x

x+ = ⇒ 22 5 2 0x x− + = ⇒

12,

2x =

13x

x+ = − ⇒ 2 3 1 0x x+ + = ⇒

3 5

2x

− ±=

Thus, 4 3 22 11 2 0x x x x+ − + + = has four roots 2, 1

2, 3 5

2

− + , 3 5

2

− −

THEORY



� ( )( )( )( )− − − − = 2 & (ab = cd)x a x b x c x d Ax

2Multiply the factors for which product of constant terms are equal and divide byx

eg. : ( ) ( ) ( ) ( ) 21 2 3 6 3x x x x x+ + + + = (((( )))). .1 6 2 3====

⇒ ( )( )2 2 27 6 5 6 3x x x x x+ + + + =

⇒( ) ( )2 27 6 5 6

. 3x x x x

x x

+ + + + =

⇒6 6

7 5 3x xx x

+ + + + =

Put 6

x yx

+ =

we get ( )( )7 5 3y y+ + =

⇒ 2 12 32 3y y+ + =

⇒ ( )( )4 8 0y y+ + =

⇒ 4, 8y = − −

64x

x+ = − ⇒ 2 4 6 0x x+ + = ⇒ 2 2ix = − ±

68x

x+ = − ⇒ 2 8 6 0x x+ + = ⇒ 4 10x = − ±

Thus, ( )( )( )( ) 21 2 3 6 3x x x x x+ + + + = has four roots 2 2i− + , 2 2i− − , 4 10− + , 4 10− −

� ( ) ( )4 4x a x b k− + − =

( ) ( )put or

2 2a b x a x b

x t+ − + −

− =

eg. : ( ) ( )4 42 4 16x x− + − =

Let 2 4

32

x xx t

− + − = − =

we get ( ) ( )4 41 1 16t t+ + − =

⇒ 4 26 7 0t t+ − =



Put 2t y= ,

we get 2 6 7 0y y+ − =

( )( )1 7 0y y− + =

1y = , 7y = −

2 1t = ⇒ 1t = ± , 3 1x − = ± , x ==== 4, 2

2 7t = − ⇒ 7it = ± , 3 7ix − = ± , 3 7ix = ±

Thus, ( ) ( )4 42 4 16x x− + − = has four roots 2, 4, 3 7i± .

2.8.2 Irrational Functions

� If roots are even (i.e., 64, , ) then 2n y implies two conditions, i.e., 0y ≥ and 0y ≥ . This should

be kept in mind before solving.

� If roots are odd (i.e., 3 5, ) then there is no such restriction.

Square (or cube or take higher power), one or more time to forma quadratic.

eg.: (i) 4 5x − = − → (no solution)

(ii) ( ) ( )1/2 1/52 7x x− − = − → (no solution)

(iii) ( )26 4 4x x x− − = +

⇒ 2

4 0 4

6 4 0

x x

x x

+ > ⇒ > − − − >

Two conditions that should be kept in mind because of

Squaring 2 26 4 8 16x x x x− − = + +

⇒ 22 12 10 0x x+ + =

⇒ 2 6 5 0x x+ + =

⇒ ( )( )5 1 0x x+ + = ⇒ 5, 1x x= − = −

5x ≠ − as 4 0x + > (from Ist condition)

Thus only solution is x = – 1 (as it satisfies both the conditions)

(iv)1/3 1/3

1 11

2 2x x + − − =

(for cube root, we can cube directly)

THEORY



⇒

2/3 1/3 1/3 2/31 1 1 1 1 1

3 3 12 2 2 2 2 2

x x x x x x + − + − + + − − − =

⇒1/3 1/3 1/3 1/3

1 1 1 13 0

2 2 2 2x x x x

+ − − − + =

1/31

02

x + =

⇒1

2x

−=

1/31

02

x − =

⇒1

2x =

1/3 1/31 1

02 2

x x − − + =

⇒ No solution (why .........?)

Thus real solutions of x are 1

2± .


1. The set of values of p for which the roots of the equation 23 2 ( 1) 0x x p p+ + − = are of opposite signs is

(a) ( ,0)−∞ (b) (0, 1) (c) (1, )∞ (d) (0, )∞

2. If 2 ( 3) 0x a x a− − + = has atleast one positive root then :

(a) ( ,0) [7,9]a∈ −∞ ∪ (b) ( ,0) [7, ]−∞ ∪ ∞

(c) ( ,0) [9, )a∈ −∞ ∪ ∞ (d) None of these

3. If the roots of 2 ( 3) 0x a x a− − + = are such that both of them are greater than 2, then

(a) [7,9]a∈ (b) [7, )a∈ ∞ (c) [9,10)a∈ (d) [7,9)

4. The equation 2 2 0x ax b+ + = has two roots each of which exceeds a number c, then

(a) 2 24a b< (b) 2 2 0c ac b+ + > (c)2

ac

− < (d) None of these

5. If the equation 2 2( 1) 9 5 0x k x k+ + + − = has only negative roots, then :

(a) 0k ≤ (b) 0k ≥ (c) 6k ≥ (d) 6k ≤

6. If a and b ( 0≠ ) are the roots of the equation 2 0x ax b+ + = , then the least value of 2 ( )x ax b x R+ + ∈ is :

(a) 9/4 (b) –9/4(c) –1/4(d) 1/4

7. If x R∈ , the least value of the expression 2

2

6 5

2 1

x x

x x

− ++ +

is :



(a) –1 (b) –1/2(c) –1/3(d) None of these

8. The number of values of k for which 2 2(16 12 39) (9 2 11)x x k x x+ + + − + is perfect square is :

(a) two (b) zero(c) one (d) None of these

9. Let ( 1)( 5)

.3

x xy

x

+ −=−

Find all x R∈ for which y is a real function of x.

10. Solve : 2

2

5 41

4

x x

x

− + ≤−

. THEORY



Solved Examples

Example 1:

If α, β be the roots of the equation 2 0x px q− + = and 0, 0α > β > , then find the value of

(i) 5 5α + β (ii) 1/4 1/4α + β

Solution :

Given equation is

2 0x px q− + = ..... (1)

has roots α and β

then pα + β = ..... (2)

and qαβ = ..... (3)

(i) ( )5 5 4 3 2 3 4 55 10 10 52α + β = α + α β + α β + α β + αβ + β (By Binomial theorem or by actual multiplication)

( ) ( ) ( )5 5 3 3 2 25 10= α + β + αβ α + β + α β α + β

( ) ( ) ( ){ } ( )35 5 2 25 3 10= α + β + αβ α + β − αβ α + β + α β α + β

∴ ( ) ( ) ( ) ( )5 35 5 2 2 2 25 15 10α + β = α + β − αβ α + β + α β α + β − α β α + β

5 3 2 25 15 10p qp q p q p= − + − {from (2) and (3)}

5 3 25 5p qp q p= − +

(ii) ( )41/4 1/4 1/4 4 1/4 3 1/4 1/4 2 1/4 1/4 1/4 3 1/4 4( ) 4( ) ( ) 6( ) ( ) 4( )( ) ( )2α + β = α + α β + α β + α β + β

(By Binomial theorem or by actual multiplication)

3/4 1/4 1/2 1/2 1/4 3/43 6 4= α + α β + α β + α β + β

1/4 1/4 1/2 1/2 1/2 1/2( ) 4 ( ) 6= α + β + α β α + β + α β

1/4 1/2 1/2 2 1/2( ) 4( ) ( ) 6( )= α + β + αβ α + β + αβ

1/4 1/2 1/2 1/2( ) 4( ) ( 2 ) 6( )= α + β + αβ α + β + α β + αβ

1/4 1/2 1/2( ) 4( ) ( 2( ) ) 6( )= α + β + αβ α + β + αβ + αβ

1/44 ( 2 ) 6p q p q q= + + +

∴ 1/4

1/4 1/4 1/44 ( 2 ) 6p q p q qα + β = + + +



Example 2:

If the equation 2 2 0ax bx c+ + = has real roots, a, b and c being real numbers and if m and n are real numbers such that2 0m n> > then prove that the equation 2 2 0ax mbx nc+ + = has real roots.

Solution :

Since roots of the equation

2 2 0ax bx c+ + = are real

∴ 2(2 ) 4 0b ac− ≥

∴ 2 0b ac− ≥ ..... (1)

and discriminant of 2 2 0ax mbx nc+ + =

is 2(2 ) 4D mb anc= −

2 24 4D m b anc= − ..... (2)

From (1) 2b ac≥ ..... (3)

and given 2m n> ..... (4)

from (3) and (4)

∴ 2 2b m anc≥

⇒ 2 24 4 0b m anc− ≥

⇒ 0D ≥ {from (2)}

Hence roots of equation 2 2 0ax mbx nc+ + = are real.

Example 3:

Find a quadratic equation whose roots 1x and 2x satisfy the condition

2 2 5 5 3 3

1 2 1 2 1 25,3( ) 11( )x x x x x x+ = + = + . (Assume that 1x and 2x are real)

Solution :

We have 5 5 3 3

1 2 1 23( ) 11( )x x x x+ = +

⇒

5 5

1 2

3 3

1 2

11

3

x x

x x

+=

+

⇒2 2 3 3 2 2

1 2 1 2 1 2 1 2

3 3

1 2

( )( ) ( ) 11

( ) 3

x x x x x x x x

x x

+ + − +=

+

⇒

2 2

2 2 1 2 1 2

1 2 2 2

1 2 1 2 1 2

( ) 11( )

( )( ) 3

x x x xx x

x x x x x x

++ − =

+ + −

⇒

2 2

1 2

1 2

115

5 3

x x

x x− =

−

SOLVED

EXAMPLES



⇒

2 21 2

1 2

4

3 5

x x

x x=

−

⇒ 2 21 2 1 23 20 0x x x x+ − =

⇒ 2 2

1 2 1 2 1 23 10 6 20 0x x x x x x+ − − =

⇒1 2 1 2( 2)(3 10) 0x x x x− + =

∴ 1 2

102,

3x x = −

We have 2 2 21 2 1 2 1 2( ) 2x x x x x x+ = + +

1 2

5 2x x= +

∴ 2

1 2( ) 5 4x x+ = + {if 1 2

2x x = }

= 9

∴1 2

3x x+ = ±

∴ 2

1 2( ) 5 2( 10/3)x x+ = + − {if 1 2

10 / 3x x = − }

which is not possible 1 2,x x are real

Thus required equations are

2 3 2 0x x± + =

Example 4:

If α, β are the roots of 2 0,x px q+ + = and ,γ δ are the roots of 2 0x rx s+ + = , evaluate

( )( )( )( )α − γ α − δ β − γ β − δ in terms of p, q, r and s. Deduce the condition that the equations have a common root.

Solution :

∴ α, β are the roots of 2 0x px q+ + =

∴ ,p qα + β = − αβ = ..... (1)

and γ, δ are the roots of 2 0x rx s+ + =

∴ ,r sγ + δ = − γδ = ..... (2)

Now, ( )( )( )( )α − γ α − δ β − γ β − δ

[ ][ ]2 2( ) ( )= α − α γ + δ + γδ β − β γ + δ + γδ

2 2( )( )r s r s= α + α + β + β + {from (2)}

2 2 2 2 2 2( ) ( ) ( )r r s sr s= α β + αβ α + β + αβ + α + β + α + β +

2 2 2 2 2( ) {( ) 2 } ( )r r s sr s= α β + αβ α+β + αβ+ α+β − αβ + α+β +

2 2 2 2( 2 ) ( )q pqr r q s p q sr p s= − + + − + − +



2 2 2( )q s rpq r q sp prs= − − + + −

2( ) ( ) ( )q s rq p r sp p r= − − − + −

2( ) ( )( )q s p r sp rq= − + − − ..... (3)

For a common root (Let α = γ or β = δ )

then ( )( )( )( ) 0α − γ α − δ β − γ β − δ = ..... (4)

From (3) & (4), we get

2( ) ( )( ) 0q s p r sp rq− + − − =

⇒ 2( ) ( )( ),q s p r qp sp− = − − which is the required condition.

Example 5:

Show that the roots of the equation 4 4 2 4 4( ) 4 0p q x pqrsx r s+ + + + = , can not be different, if real.

Solution :

The discriminant of the given equation

2 2 2 2 4 4 4 416 4( )( )D p q r s p q r s= − + +

4 4 4 4 2 2 2 24 ( )( ) 4p q r s p q r s= − + + −

4 4 4 4 4 4 4 4 2 2 2 24 4p r p s q r q s p q r s= − + + + −

2 2 2 2 2 2 2 2 2 24 ( ) ( )p r q s p s q r= − − + −

If roots of the given equation are real then 0D ≥

or 2 2 2 2 2 2 2 2 2 24 ( ) ( ) 0p r q s p s q r− − + − ≥

⇒2 2 2 2 2 2 2 2 2 24 ( ) ( ) 0p r q s p s q r− + − ≤

But 2 2 2 2 2 2 2 2 2 24 ( ) ( ) 0p r q s p s q r− + − </

∴ 2 2 2 2 2 2 2 2 2 24 ( ) ( ) 0p r q s p s q r− + − =

∴ from (1), D = 0

Hence roots of the given equation are equal.

Example 6:

If each pair of the following three equations 2 2 20, 0, 0x ax b x cx d x ex f+ + = + + = + + =

has exactly one root in common, then show that 2( ) 4( )a c e ac ce ea b d f+ + = + + − − − .

Solution :

Given equations are

2 0x ax b+ + = ..... (1)

2 0x cx d+ + = ..... (2)

SOLVED

EXAMPLES



2 0x ex f+ + = ..... (3)

Let α, β be the roots of (1), β, γ be the roots of (2) and γ, δ be the roots of (3).

∴ ,a bα + β = − αβ = ..... (4)

,c dβ + γ = − βγ = ..... (5)

,e fγ + α = − γα = ..... (6)

∴ L.H.S. 2( )a c e= + +

2( )= −α − β − β − γ − γ − α {from (4), (5), (6)}

24( )= α + β + γ ..... (7)

R.H.S. ( )ac ce ea b d f= + + − − −

{ }4 ( )( ) ( )( ) ( )( )= α + β β + γ + β + γ γ + α + γ + α α + β − αβ − βγ − γα {from (4), (5), (6)}

( )2 2 24 2 2 2= α + β + γ + αβ + βγ + γα

24( )= α + β + γ ..... (8)

From (7) & (8),

2( ) 4( )a c e ac ce ea b d f+ + = + + − − −

Example 7:

Find all values of 'a' for which the equation ( ) ( )22 2 2 1 1 0x a x a a− + + − = has roots α and β satisfying the

condition α < a < β.

Solution :

Q Number 'a' lies between the roots of the given equation then

2 ( ) 0f a <

⇒ ( ) 0f a <

∴ 22 2(2 1) ( 1) 0a a a a a− + + − <

⇒ 2 3 0a a− − < α βa

⇒ ( 3) 0a a+ >

From wavy curve method.

( , 3) (0, )a∈ −∞ − ∪ ∞

Example 8:

Solve the inequation 72 1 13 1

3 3

x x

>

.

Solution :

The given inequation is valid only when 0x ≥The given inequation can be written in the form



723 1x x− − > .

⇒ 72 0x x− − > (Q 3 > 1)

⇒ 72 0x x+ − <

⇒ ( 9)( 8) 0x x+ − <

But 9 0x + > for all 0x ≥

∴ 8 0x − <

⇒ 8x <∴ 0 64x≤ < {from (1)}

Thus the interval [0,64)x∈ is the set of all solutions of the given inequation.

Example 9:

Find the real roots of the equation 2 2 ...... 2 2 3x x x x x x+ + + + + =( radical signs)n

Solution :

Rewrite the given equation

2 2 ...... 2 2 3x x x x x x+ + + + + =

On replacing the last letter x on the L.H.S. of equation (1) by the value of expressed by (1) we obtain.

2 2 ...... 2x x x x x x= + + + + +(2 radical signs)n

Further, let us replace the last letter x by the same expression; again and again yields

∴ 2 2 ...... 2x x x x x x= + + + + +(3 radical signs)n

2 2 ...... 2 2 ...x x x x x= + + + + + =(4 radical signs)n

We can write

2 2 ...x x x x= + + + =

lim 2 2 ...... 2 2N

x x x x x→∞

= + + + + +( radical signs)N

It follows

2 2 ...x x x x= + + + =

( )2 2 ... 2x x x x x= + + + = +

SOLVED

EXAMPLES



Hence 2 22 3 0x x x x x= + ⇒ − =∴ 0,3x = .

Example 10:

If 3 22 4 0x ax bx+ + + = (a and b are positive real numbers) has 3 real roots, then prove that1/312(2 )a b+ > .

Solution :

Let α, β, γ be the roots of the given equation. Then,

2

aα + β + γ = − ..... (1)

2

bαβ + βγ + γα = ..... (2)

and 2αβγ = − ..... (3)

Since all the coefficients of the given equation are positive, so its all roots are negative. Let 1 1,α = −α β = −β and

1γ = −γ

∴ from (1), (2) and (3), we obtain

1 1 12

aα + β + γ = ..... (4)

1 1 1 1 1 12

bα β + β γ + γ α = ..... (5)

and1 1 1

2α β γ = ..... (6)

Now A.M. G.M.≥

⇒1/31 1 1

1 1 1( )3

α + β + γ≥ α β γ

⇒1/3 1/32 6(2 )

6

aa≥ ⇒ ≥ ..... (7)

Again A.M. G.M.≥

⇒ { }1/31 1 1 1 1 11 1 1 1 1 1( )( )( )

3

α β + β γ + γ α≥ α β β γ γ α

⇒ 1/36(2 )b ≥ ..... (8)

From (7) and (8), we obtain

1/3 2/36(2 2 )a b+ ≥ +

⇒1/3 1/36(2 2 )a b+ > +

⇒ 1/312(2 )a b+ >



Exercises

Level - 1

Single Choice Questions

1. If 2 0x ax b+ + = and 2 0;( )x bx a a b+ + = ≠ have a common root, the numerical value of a + b is(a) 1 (b) –1 (c) ± 1 (d) None of these

2. If one root of 2 2 3 0x x k+ + = and 22 3 5 0x x k+ + = is common, then k is

(a) 0 (b) –1 (c) 0 or –1 (d) None of these

3. If the equations 2 0ax bx c+ + = and 2 0cx bx a+ + = ( )a c≠ have a negative common root, then the value of (a – b

+ c) is

(a) 0 (b) 2 (c) 1 (d) None of these

4. If 2 21 0x hx− − = and 2 3 35 0x hx− + = (h > 0) have a common root, then h is

(a) 2 (b) 4 (c) 3 (d) 35/21

5. The equation 2 2 1 2x b bx+ = − and 2 2 1 2x a ax+ = − have one and only one root common. Then

(a) 2a b− ≠ (b) 2 0a b− + ≠ (c) | | 2a b− ≠ (d) None of these

6. The equation 2 3 20, 2 2 1 0ax bx a x x x+ + = − + − = have two roots common. Then a + b must be equal to

(a) 1 (b) –1 (c) 0 (d) None of these

7. If 2 21 0x ax− − = and 2 3 35 0x ax− + = , a > 0 have a common root, then a is

(a) – 4 (b) 4 (c) 2 (d) None of these

8. If , , ,p q r s R∈ and 2( )pr q s= + and 2 20, 0x px q x rx s+ + = + + = are two equations then which of the following

statement is true ?

(a) both the equation have real equal roots (b) both the equations have real and distinct roots

(c) at least one of the equations has real roots (d) None of these

9. If , ,a b c R∈ and 1 is a root of the equation 2 0ax bx c+ + = , the equation 24 3 2 0, 0ax bx c c+ + = ≠ has

(a) imaginary roots (b) real and equal roots

(c) real and unequal roots (d) rational roots

10. Let α ≠ β and 2 2 5α + = α and 2 5 3β = β − . The quadratic equation whose roots are α/β and β/α is

(a) 23 31 3 0x x− + = (b) 23 19 3 0x x− + = (c) 23 19 3 0x x+ + = (d) None of these

Multiple Choice Questions

11. If 1 2 3 4 5 6,a a a a a a< < < < < then the equation 1 2 3 4 5 6( )( )( )( )( )( ) 0x a x a x a x a x a x a− − − − − − = has

(a) three real roots (b) a root in 1( , )a−∞ (c) one root in 1 2( , )a a (d) one root in 5 6( , )a a

12. 0 < a < b < c and the roots a, b of the equation 2 0ax bx c+ + = are imaginary, then

(a) | | | |α = β (b) | | 1α > (c) | | 1β < (d) None of these

SOLVED

EXAMPLES



13. If ( )f x is divided by ( 1), ( 1)x x− + gives remainder 4 and 8 respectively, the remainder when f(x) is divided by

( 1) ( 1)x x− + is

(a) 2 6x− + (b) 6x + (c) 6x − (d) 12x −

14. The remainder when 3 3x px q+ + is divided by 2( )x a− , is

(a) 2 33 2x q a+ − (b) 2 33( ) 2p a x q a+ + − (c) 32q a− (d) None of these

15. If one root of 2 14 8 0kx x− + = may be six times the other. then, k is


Comprehension Type

Comprehension 1

Let 2y ax bx c= + + be a quadratic expression having its vertex at (3, –2) and value of c = 10, then :

Answer the following questions :

16. Value of b is equal to

(a) 6 (b) –6 (c) 8 (d) –8

17. One of the roots of the equation 2 0ax bx c+ + = is

(a)6 6

2+

(b)3 6

2+

(c) 3 6− (d) 3 6+

18. If 23

y ≥ − , then

(a) ( ,2] [4, )x ∈ −∞ ∪ ∞ (b) ( ,3] [4, )x ∈ −∞ ∪ ∞ (c) ( ,1] [3, )x ∈ −∞ ∪ ∞ (d) ( ,4] [6, )x ∈ −∞ ∪ ∞

19. If the equation ( ) 0f x = has roots α and β, then the equation whose roots are α2 and β2 is

(a) 2 24 30 0x x− − = (b) 2 24 30 0x x+ − = (c) 24 44 225 0x x− + = (d) 24 84 225 0x x− + =

20. The graph of | ( ) |y f x= is

(a) (b)

(c) (d) None of these

Comprehension 2

Let 2 21 1 2 2( ) , ( )f x x b x c g x x b x c= + + = + + real roots of ( ) 0f x = be a, b and real roots of ( ) 0g x = be ,α + δ β + δ . Least

value of ( )f x be 14

− . Least value of ( )g x occurs at 72

x = .




21. The least value of g(x) is

(a) 1 (b)12

− (c)14

− (d)13

−

22. The value of b2 is

(a) 6 (b) –7 (c) 8 (d) 0

23. The roots of ( ) 0g x = are

(a) 3, 4 (b) –3, 4 (c) 3, –4 (d) –3, –4

Comprehension 3

Let 2( ) 0f x ax bx c= + + = be a quadratic expression and ( )y f x= has graph as shown in figure.


24. Which of the following is false

(a) ab > 0 (b) abc < 0 (c) ac < 0 (d) bc < 0

25. Which of the following is true

(a) 0a b c+ + > (b) 0a b c− + > (c) 3 9 0a b c+ + < (d) 3 9 0a b c− + >

26. If ( )f x is an integer whenever x is an integer, then which of the following is always correct ?

(a) a is an integer (b) 2a is an integer (c) b is an integer (d)c/2 is an integer

Matrix Match

27. Match the Columns

Column - I Column - II

(A) 2 2( 1) (9 5) 0x k x k+ + + − = has (p) 2 < k < 4

only negative roots

(B) 2 22(4 1) 15 2 7 0x k x k k− − + − − > for all k (q) 6k ≥

(C) 2 2( 1) (2 1) 0x k x k− − + + = both positive roots (r) 1k < − or k > 0

(D) 22 2(2 1) ( 1) 0x k x k k− + + + = one root less (s) 4k ≥

then k, other greater than k.

(E) graph of 2 3 2x x y= − − is strictly below y k= (e)14

k =

EXERCISES



Assertion and Reason Type

Some questions (Assertion-Reason type) are given below. Each question contains Statement I (Assertion) and Statement II(Reason). Each question has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct. So, select the correct choice.(a) Statement I is True, Statement II is True; Statement II is a correct explanation for Statement I(b) Statement I is True, Statement II is True; Statement II is NOT a correct explanation for Statement I(c) Statement I is True, Statement II is False(d) Statement I is False, Statement II is True

28. Statement I : If 2, 2 3 5x R x x∈ + + is positive.

Statement II : If 20, ,ax bx c∆ < + + ‘a’ have same sign x R∀ ∈

29. Statement I : If 1 2+ is a root of 2 2 1 0x x− − = , then 1 2− will be the other root.Statement II : Irrational roots of a quadratic equation with rational coefficients always occur in conjugate

pair.

Level - 2

Single Choice Questions

1. If α be a root of the equation 24 2 1 0x x+ − = , then

(a) 34 3α + α is the other root (b) 34 3α − α is the other root

(c) 24 2α + α is the other root (d) 24 2α − α is the other root

2. If the quadratic equation 2 2 0ax cx b+ + = and 2 2 0,( )ax bx c b c+ + = ≠ have a common root. Then 4 4a b c+ + is

equal to

(a) 0 (b) –1/2 (c) 2 (d) 1

3. If 2 1 0ax mx+ + = and 2( ) ( ) ( ) 0a b x b c x c a− + − + − = have both common root, then which of the following is false ?

(a) m = – 2 (b) b, a, c are in A.P. (c) 2 3 1a b c− + = (d) a2 + a - ac = b

4. If equation 2 0ax bx c+ + = and 2 2 3 0x x+ + = have a common root, then a : b : c is

(a) 1 : 2 : 3 (b) 3 : 2 : 1 (c) 4 : 9 : 16 (d) None of these

5. If the equation 2 0ax bx c+ + = and 3 23 3 2 0x x x+ + + = have two common roots, then

(a) a b c= ≠ (b) a b c= − = (c) a b c= = (d) None of these

6. The value of 'a' for which the equation 3 1 0x ax+ + = and 4 2 1 0x ax+ + = have a common root is

(a) 2 (b) –2 (c) 0 (d) None of these

7. If every pair from among the equation 2 20, 0x px qr x qx rp+ + = + + = and 2 0,x rx pq+ + = have a common root,

then sum of roots

product of roots

is

(a)p

pqrΣ

(b)1pq

Σ (c) 2( )p q r+ + (d) None of these



Multiple Choice Questions

8. If 3 2( ) 2f x x x k= + + is divisible by ( 1)x− , then the value of k is

(a) –3 (b) –2 (c) –1 (d) 0

9. If α β, are the roots of the equation 2 1 0x x+ − = , then the equation whose roots are 2,α β+ + 2 is

(a) 2 1 0x x+ + = (b) 2 1 0x x− + = (c) 2 3 1 0x x− + = (d) 2 3 1 0x x− − =

10. If 2 3+ is one root of 2 0x px q+ + = , then p and q are

(a) –4, –1 (b) 4, –1 (c) –4, 1 (d) 4, 1

11. The number of solutions for 5 7 3 1 3x x x+ − + = + is


12. The value of ‘a’ for which the equation, 2 220 ( 4 ) 0x x a a− + − = has roots of opposite sign, is

(a) 4a > (b) 4a < (c) 0 4a< < (d) 0a <

Comprehension Type

Comprehension 1

If 2 3x i= + is a root of 2 0x px q+ + = when p, q are real.


13. Value of p is

(a) – 3 (b) – 4 (c) 4 (d) 3

14. Value of q is

(a) 2 (b) 7 (c) 3 (d) 4

15. Graph of 2y x px q= + + is

(a) (b)

(c) (d)

Comprehension 2

If the only root of the equation 4 3 212 81 0x x bx cx− + + + = are positive, then

EXERCISES




16. The value of b is

(a) – 54 (b) 54 (c) 27 (d) – 27

17. The value of c is

(a) 108 (b) – 108 (c) 54 (d) – 54

18. The roots of equation 2 0bx c+ = are

(a)12

− (b)12

(c) 1 (d) – 1

Matrix Match

19. Match the Columns

Column - I Column - II

(A) If ( ) ( )

log log

x y z x y z x y

x y

+ − + −= (p) 1

( )

logy x z yz x y z

and a x y b y zz

+ −= ⋅ ⋅ = ⋅ ⋅ x yc z z= ⋅ ⋅ then a b

c

+ equals

(B) 10 10 10

1 1 1

1 log 1 log log10 , 10 implies 10 , thenx y a b zy z x a b− − += = = − equals (q) 2

(C) If 2 2 2a b c+ = log log log log ,b c c b c b c ba a k a a+ − + −⇒ + = ⋅ (r) 3

then k equals

(D) If , where 0, 0, 0 and , , 1b ac a b c a b c= > > > ≠ and if (s) 0

log log

log ,log log

a ba

b c

N Nk c

N N

−=

− then k equals

Assertion and Reason Type

Some questions (Assertion-Reason type) are given below. Each question contains Statement I (Assertion) and Statement II(Reason). Each question has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct. So, select the correct choice.(a) Statement I is True, Statement II is True; Statement II is a correct explanation for Statement I(b) Statement I is True, Statement II is True; Statement II is NOT a correct explanation for Statement I(c) Statement I is True, Statement II is False(d) Statement I is False, Statement II is True

20. Statement I : The roots of the equation 22 3 2 0x ix+ + = are always conjugate pair..Statement II : Imaginary roots of a quadratic equation with real coefficients always occur in conjugate pair.

21. Statement I : The number of real roots of 2| | | | 2 0x x+ + = is zero.

Statement II : , | | 0x R x∀ ∈ ≥ .



Previous Years’ IITJEE Questions

1. Find the all integral values of x for which 2 19 92x x+ + is perfect square.

2. Find greatest value of 2 2

12 5ax x a− − in [–3, 5] depending upon 'a'.

3. 29x a x+ < − . Find 'a', so that the given inequation has at least one negative solution.

4. ( )( )3 2 2 23 18 32 18 32 4x x x x x x= + + − − −

5. a, b, c be sides of ∆, None of them are equal and Rλ ∈ , if roots of equation

2 2( ) 3 ( ) 0x a b c x ab bc ca+ + + + λ + + = are real then

(a)43

λ < (b)53

λ > − (c)1 5

,3 3

λ ∈

(d)4 5

,3 3

λ ∈

6. If α, β are the roots of 2 0ax bx c+ + = and n nns = α + β , show that 1 1 0n n nas bs cs+ −+ + = . Hence find 5s .

7. Find integral values of 'x' for which 2 (5 ) 5 7 0x a x a− + + − = for integral values of 'a'.

8. If equations 3 23 3 0x px qx r+ + + = and 2 2 0x px q+ + = have a common root, show that

2 2 24( )( ) ( )p q q pr pq r− − = − .

9. If each pair of the following three equation 2 2 20, 0, 0x ax b x cx d x ex f+ + = + + = + + = has exactly one

root in common, then show that 2( ) 4( ).a c e ac ce ea b d f+ + = + + − − −

10. Find all values of 'a' for which the equation 3 2 2 1 1a a a x a x+ + + + = has not less than 4 different integer

solutions.



Answer Key


1. (b) 2. (b) 3. (a)


1. (b) 2. (c) 3. (c) 4. (b) 5. (c) 6. (b)

7. (b) 8. (a) 9. [ 1, 3) [5, )− ∪ ∞ 10.5 5

0, ,8 2

∪ ∞

Level - 1

1. (b) 2. (c) 3. (a) 4. (b) 5. (d) 6. (c)

7. (b) 8. (c) 9. (c) 10. (b) 11. (a) 12. (a, b)

13. (a) 14. (b) 15. (b) 16. (d) 17. (a) 18. (a)

19. (d) 20. (d) 21. (c) 22. (b) 23. (a) 24. (b)

25. (d) 26. (b) 27. (A) - p, q, s; (B) - p; (C) - p, q, s; (D) - r; (E) - t

28. (a) 29. (a)

Level - 2

1. (b) 2. (a) 3. ( d) 4. (a) 5. (c) 6. (b)

7. (a) 8. (a) 9. (c) 10. (d) 11. (b) 12. (c)

13. (b) 14. (b) 15. (b) 16. (b) 17. (b) 18. (c)

19. (A) - p; (B) - q; (C) - q; (D) - p 20. (d) 21. (a)

Answer Key : Previous years’ IIT JEE Questions

1. No values of x 2. 2 2

1 (5) 1; ( )

1 ( 3) 2 5

a ff x

a f ax x a

<= > − − −

3.37

9,4

a ∈ −

4. 3 7, 3 17x i= − ± ± 5. (a) 7. {11, 1}− 10. ( , 3) (0,1/ 2)a∈ −∞ − ∪

Documents

Equation and Inequation